Graphics Reference
In-Depth Information
3.5
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2.5
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1.5
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0.5
0
0
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x
Fig. 8.9. A cubic interpolant between points (1, 1) and (4, 3).
d V out
d t
The rate of change of V out relative to t (i.e.
)mustequal1when t =0,
d V out
d t
so it can be used to multiply s 1 .When t =1 ,
must equal 0 to attenuate
any trace of s 1 :
d V out
d t
=3 at 2 +2 bt + c
(8.22)
d V out
d t
=1when t =0,and d V out
d t
but
=0when t = 1. Therefore c =1,and
3 a +2 b + 1 = 0
(8.23)
But using (8.21) means that b =
2and a = 1. Therefore, the polynomial
V out has the form
V out = t 3
2 t 2 + t
(8.24)
Using a similar argument, one can prove that the function to blend in s 2 equals
V in = t 3
− t 2
(8.25)
Graphs of (8.24) and (8.25) are shown alongside graphs of (8.16) in Figure 8.10.
The complete interpolating function looks like
n 1
n 2
s 1
s 2
n =[2 t 3
3 t 2 +1
2 t 3 +3 t 2
t 3
2 t 2 + t 3
t 2 ]
·
(8.26)
 
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